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The given data can be listed below as-

The energy of the photon at the ground state is,$E_0=0.17\text{eV}$.

The energy of the photon in the ground state can be determined by taking half of the product of Planck鈥檚 constant and wavelength of the photon. It is expressed as follows,

$E_0=\left(\frac{1}{2}\right)h{蝇}_0$

Here, $h$is the Planck鈥檚 constant and ${蝇}_0$is the wavelength of the photon.

The equation of the photon鈥檚 energy can be expressed as,

$$\begin{gathered} E_n=n+E_0 \\ =n+\left(\frac{1}{2}\right)h{蝇}_0 \end{gathered}$$

Here, $n$is the number of states, $E_0$is the energy of the photon at the ground state that is half of the product of the Planck鈥檚 constant $(h)$and wavelength of the photon $\left({蝇}_0\right)$$({蝇}_0)$.

Substitute localid="1662469232427" $n=2$for the second excited state in the above expression.

localid="1662469239412" $E_2=2+$localid="1662469245507" $\left(\frac{1}{2}\right)h{蝇}_0$

localid="1662469252499" $\frac{5}{2}h{蝇}_0$

Substitute localid="1662469258038" $n=5$for the fifth excited state in the above expression.

localid="1662469265782" $$\begin{gathered} E_5=5+\left(\frac{1}{2}\right)h{蝇}_0 \\ =\left(\frac{11}{2}\right)h{蝇}_0 \end{gathered}$$

The equation of the change in the energy transition of the photon can be expressed as,

$螖E=E_5-E_2$

Here, $E_5$is the energy of the fifth excited state and $E_2$is the energy of the second excited state.

For ,$E_2=\left(\frac{5}{2}\right)h{蝇}_0$and $E_5=\left(\frac{11}{2}\right)h{蝇}_0$and $\left(\frac{1}{2}\right)h{蝇}_0=0.17eV$

$$\begin{gathered} 鈭�E=\left(\frac{11}{2}\right)h{蝇}_0-\left(\frac{5}{2}\right)h{蝇}_0 \\ =6脳\left(\frac{1}{2}\right)h{蝇}_0 \\ =6脳0.17eV \\ =1.02eV \end{gathered}$$

Thus, the energy of a photon emitted in a transition is $1.02\text{eV}$.